\newproblem{lay:5_5_7}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.5.7}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let the matrix $A=\begin{pmatrix}\sqrt{3} & -1 \\ 1 & \sqrt{3}\end{pmatrix}$. Consider the transformation $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ defined as
	$T(\mathbf{x})=A\mathbf{x}$. $T$ is the composition of a scaling and a rotation. Give the scaling factor and the rotation angle.
}{
  % Solution
	The eigenvalues of $A$ are
	\begin{center}
		$|A-\lambda I|=0$ \\
		$\left|\begin{array}{cc}\sqrt{3}-\lambda & -1 \\ 1 & \sqrt{3}-\lambda\end{array}\right|=(\sqrt{3}-\lambda)^2+1=0$\\
		$\sqrt{3}-\lambda=\pm i$ \\
		$\lambda=\sqrt{3}\pm i=2e^{\pm i 30\degree}$
	\end{center}
	The scaling factor is 2 and the rotation angle 30\degree~or -30\degree (in fact looking only at the eigenvalues we cannot determine the sign of the rotation). However
	We note that 
	\begin{center}
		$A=\begin{pmatrix}\sqrt{3} & -1 \\ 1 & \sqrt{3}\end{pmatrix}=\begin{pmatrix}2 & 0 \\ 0 & 2\end{pmatrix}
		   \begin{pmatrix}\cos(30\degree) & -\sin(30\degree) \\ \sin(30\degree) & \cos(30\degree)\end{pmatrix}$
	\end{center}
	So, the rotation angle is $30\degree$.
}
\useproblem{lay:5_5_7}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
